Question

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = xy i + yz j + zx k S is the part of the paraboloid z = 8 − x2 − y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation

Answer 1

Parameterize $S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k$

with $0\le u\le1$ and $0\le v\le1$.

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}$